Question

a 1200 kg car rolling on a horizontal surfact has a speed of 25 m/s when it strikes a horizontal coiled spring and is brought ot rest in a distance of 2.5m. what is the spring constant of this spring

Answer 1

The required spring constant:

The spring constant of the spring is $12\times 10^4 \text{ N/m}$.

Calculation:

The mass of the car is m=1200 kg, the speed of the car is v=25 m/s, and after colliding the spring is brought to rest at a distance of x=2.5m. Let the spring constant of the spring is, k.

From the conservation of energy,

Total initial kinetic energy= Total final potential energy of the spring

Therefore,

$\frac{1}{2}mv^2=\frac{1}{2}kx^2$

Now, substituting the values of the mass of the car, speed of the car, and displacement, we get:

$\begin{aligned}1200\times(25)^2&=k\times (2.5)^2\\\Rightarrow k&=\frac{1200\times(25)^2}{(2.5)^2}\\&=12\times 10^4 \text{ N/m}\end{aligned}$

To know more about spring constant, refer to:

brainly.com/question/14159361

#SPJ4